Question

If $y=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\cdots$ to $\infty$ with $|x|>1$, then $\frac{dy}{dx}=$

• A. $-\frac{y^2}{x^2}$
• B. $\frac{y^2}{x^2}$
• C. $x^2{y}^2$
• D. $\frac{x^2}{y^2}$

Answer 1

The correct option is A $-\frac{y^2}{x^2}$

The given equation is:

$y=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+.......\infty$

The series is a G.P. with infinite terms and the common ratio is also less than 1. Hence the sum of the series is given by

$\Rightarrow y=\frac{1}{1-\frac{1}{x}}$

$\Rightarrow y=\frac{x}{x-1}$

Differentiating once w.r.t. to x we get,

$\Rightarrow \frac{dy}{dx}=\frac{(x-1)-x}{(x-1{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{(x-1{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{-x^2}{x^2(x-1{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{-y^2}{x^2}$ .....Answer